Question: A bag contains $6$ red balls, $4$ green balls, and $3$ blue balls. If we choose a ball, then another ball without putting the first one back in the bag, what is the probability that the first ball will be green and the second will be red?
Explanation: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened. In this case, event A is picking a green ball and leaving it out. Event B is picking a red ball. Let's take the events one at at time. What is the probability that the first ball chosen will be green? There are $4$ green balls, and $13$ total, so the probability we will pick a green ball is $\dfrac{4} {13}$. After we take out the first ball, we don't put it back in, so there are only $12$ balls left. Since the first ball was green, there are still $6$ red balls left. So, the probability of picking a red ball after taking out a green ball is $\dfrac{6} {12}$. Therefore, the probability of picking a green ball, then a red ball is $\dfrac{4}{13} \cdot \dfrac{6}{12} = \dfrac{24}{156} = \dfrac{2}{13}$